Question

A sample of 5 hotels reported the price for a night's stay:
$189 $219 $259 $329 $129
What is the range of the prices of a night’s stay? How to find the standard deviation of the nightly prices

Answer 1

Answer:

a) Range = 200

b) Standard Deviation = 75.03

Step-by-step explanation:

Prices: $189 $219 $259 $329 $129

a) Range

Range = Maximum value - Minimum value

Maximum value = 329

Minimum value = 129

So, Range = 329 - 129

Range = 200

b) Standard Deviation

The formula used for finding standard deviation is:

$\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_{i}-\mu)^2}$

N is No of terms

μ is mean

Mean μ = (189+219+259+329+129)/5 = 225

x              x-μ                  (x-μ)^2

189      189-225= -36      1296

219      219 - 225= -6       36

259     259-225= 34       1156    

329      329-225=104      10,816

129       129-225=-96        9216

Now, find $\sum_{i=1}^N(x_{i}-\mu)^2$

$\sum_{i=1}^N(x_{i}-\mu)^2=1296+36+1156+10816+9216$

$\sum_{i=1}^N(x_{i}-\mu)^2=22520$

Now finding standard deviation $\sigma$

$\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_{i}-\mu)^2}$

$\sigma=\sqrt{\frac{1}{5-1}(22520)}\\\sigma=\sqrt{5630}\\\sigma=75.03$